当$x,y\ge0,x+y=2$时求下面式子的最小值:
1)$x+\sqrt{x^2-2x+y^2+1}$
2)$\dfrac{1}{5}x+\sqrt{x^2-2x+y^2+1}$
解:1)$P(x,y)$为直线$x+y=2$上一点,点$H$为$P$到$y$轴的投影点,
设$A(1,0)$则$A$关于$x+y=2$的对称点$A‘(2,1)$
故$x+\sqrt{x^2-2x+y^2+1}=|PH|+|PA|= |PH|+|PA‘|\ge|HA‘|=2$
2)$\dfrac{1}{5}x+\sqrt{x^2-2x+y^2+1}$
$=\dfrac{1}{5}x+\sqrt{(x^2-2x+y^2+1)(\cos^2\theta+\sin^2\theta)}$
$\ge(\dfrac{1}{5}+\cos\theta)x+y\sin\theta-\cos\theta$
令$\cos\theta=\dfrac{3}{5},\sin\theta=\dfrac{4}{5}$则最小值为1
原文：https://www.cnblogs.com/mathstudy/p/10692188.html